The solution sets of homogeneous linear systems provide an important source of vector spaces. Let *A* be an *m* by *n* matrix, and consider the homogeneous system

Since *A* is *m* by *n*, the set of all vectors **x** which satisfy this equation forms a subset of **R** ^{n }. (This subset is nonempty, since it clearly contains the zero vector: **x** = **0** always satisfies *A* **x** = **0**.) This subset actually forms a subspace of **R** ^{n }, called the **nullspace** of the matrix *A* and denoted *N(A)*. To prove that *N(A)* is a subspace of **R** ^{n }, closure under both addition and scalar multiplication must be established. If **x** _{1} and **x** _{2} are in *N(A)*, then, by definition, *A* **x** _{1} = **0** and *A* **x** _{2} = **0**. Adding these equations yields

which verifies closure under addition. Next, if **x** is in *N(A)*, then *A* **x** = **0**, so if *k* is any scalar,

verifying closure under scalar multiplication. Thus, the solution set of a homogeneous linear system forms a vector space. Note carefully that if the system is *not* homogeneous, then the set of solutions is *not* a vector space since the set will not contain the zero vector.

**Example 1**: The plane *P* in Example 7, given by 2 *x* + *y* − 3 *z* = 0, was shown to be a subspace of **R** ^{3}. Another proof that this defines a subspace of **R** ^{3} follows from the observation that 2 *x* + *y* − 3 *z* = 0 is equivalent to the homogeneous system

where *A* is the 1 x 3 matrix [2 1 −3]. *P* is the nullspace of *A*.

**Example 2**: The set of solutions of the homogeneous system

forms a subspace of **R**^{n} for some *n*. State the value of *n* and explicitly determine this subspace.

Since the coefficient matrix is 2 by 4, **x** must be a 4‐vector. Thus, *n* = 4: The nullspace of this matrix is a subspace of **R**^{4}. To determine this subspace, the equation is solved by first row‐reducing the given matrix:

Therefore, the system is equivalent to

that is,

If you let *x* _{3} and *x* _{4} be free variables, the second equation directly above implies

Substituting this result into the other equation determines *x* _{1}:

Therefore, the set of solutions of the given homogeneous system can be written as

which is a subspace of **R** ^{4}. This is the nullspace of the matrix

**Example 3**: Find the nullspace of the matrix

By definition, the nullspace of *A* consists of all vectors **x** such that *A* **x** = **0**. Perform the following elementary row operations on *A*,

to conclude that *A* **x** = **0** is equivalent to the simpler system

The second row implies that *x* _{2} = 0, and back‐substituting this into the first row implies that *x* _{1} = 0 also. Since the only solution of *A* **x** = **0** is **x** = **0**, the nullspace of *A* consists of the zero vector alone. This subspace, { **0**}, is called the **trivial subspace** (of **R** ^{2}).

**Example 4**: Find the nullspace of the matrix

To solve *B* **x** = **0**, begin by row‐reducing *B*:

The system *B* **x** = **0** is therefore equivalent to the simpler system

Since the bottom row of this coefficient matrix contains only zeros, *x* _{2} can be taken as a free variable. The first row then gives

satisfies *B* **x** = **0**. The collection of all such vectors is the nullspace of *B*, a subspace of **R** ^{2}: